I've been trying to show that $\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=\frac{3\pi}{8a^5}$ for $a>0$ using complex analysis methods. But for some reason I can't get it to come out. Perhaps someone could figure out where I am going wrong.

Since there are no poles on the real axis, I know that $\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=2\pi i\cdot\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right).$

To calculate $\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)$ I used the fact that on a small enough disk centered at $ai$, $\frac{1}{(z+ai)^3}=\sum\limits_{k=0}^\infty c_k(z-ai)^k$. Thus $\frac{1}{(z^2+a^2)^3}=\frac{\frac{1}{(z+ai)^3}}{(z-ai)^3}=\sum\limits_{k=0}^\infty c_k(z-ai)^{k-3}$. So $\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=c_2.$ Where $c_2=\frac{d^2}{dz^2}\frac{1}{(z+ai)^3}\bigg|_{z=ai}=\frac{12}{(2ia)^5}=\frac{3}{8ia^5}.$ But that gives me $\int_{-\infty}^\infty\frac{1}{(x^2+a^2)^3}dx=2\pi i\cdot\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=2\pi i\cdot\frac{3}{8ia^5}=\frac{3\pi}{4a^5}.$ Which is off by $\frac{1}{2}$. I must be making a silly mistake somewhere, but I can't seem to find it.

Any help would be appreciated.


There is a mistake in your calculus of $c_2$, it is rather

$$\text{Res}\left(\frac{1}{(z^2+a^2)^3},ia\right)=\color{red}{\frac{1}{2!}}\frac{d^2}{dz^2}\frac{1}{(z+ai)^3}\bigg|_{z=ai}$$ due to the Taylor expansion around $z=ia$.

  • $\begingroup$ Ooh! Gah. Of course. I'm always forgetting the factorial... Thanks so much! $\endgroup$ – mi986 May 10 '15 at 11:22
  • $\begingroup$ You are welcome! $\endgroup$ – Olivier Oloa May 10 '15 at 11:23

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